# Question f372a

Apr 17, 2017

$\textsf{p H = 8.86}$

#### Explanation:

Sodium acetate is the salt of a strong base and a weak acid so we would expect it to be slightly alkali due to salt hydrolysis:

$\textsf{C {H}_{3} C O {O}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s C {H}_{3} C O O H + O {H}^{-}}$

For which:

$\textsf{{K}_{b} = \frac{\left[C {H}_{3} C O O H\right] \left[O {H}^{-}\right]}{\left[C {H}_{3} C O {O}^{-}\right]}}$

Before we set up an ICE table we need to find its initial concentration.

The number of moles is given by:

$\textsf{{n}_{C {H}_{3} C O O N a} = \frac{m}{M} _ r = \frac{4.0}{82.034} = 0.0487}$

We get the initial concentration using $\textsf{c = \frac{n}{v}}$:

$\textsf{\left[C {H}_{3} C O O N a\right] = \frac{0.0487}{0.500} = 0.09752 \textcolor{w h i t e}{x} \text{mol/l}}$

Now set up the ICE table based on $\textsf{\text{mol/l}}$:

$\textsf{\text{ } C {H}_{3} C O {O}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s C {H}_{3} C O O H + O {H}^{-}}$

$\textsf{I \text{ "0.09752" "0" } 0}$

$\textsf{C \text{ "-x" "+x" } + x}$

$\textsf{E \text{ "(0.09752-x)" "x" } x}$

$\therefore$$\textsf{{K}_{b} = \frac{{x}^{2}}{\left(0.09752\right) - x}}$

Since we are given $\textsf{p {K}_{a}}$ we use this expression:

$\textsf{p {K}_{a} + p {K}_{b} = 14}$

$\therefore$$\textsf{p {K}_{b} = 14 - 4.74 = 9.26}$

This means $\textsf{{K}_{b}}$ is a very small number, so small that we can make the approximation that $\textsf{\left(0.09752 - x\right) \Rightarrow 0.09752}$

So we can write:

$\textsf{{K}_{b} = {x}^{2} / 0.09752}$

$\therefore$$\textsf{{x}^{2} = {K}_{b} \times 0.09752}$

sf(x=sqrt(K_bxx0.09752)#

Since $\textsf{x = \left[O {H}^{-}\right]}$ we can take negative logs of both sides to get:

$\textsf{p O H = \frac{1}{2} \left[p {K}_{b} - \log \left(0.09752\right)\right]}$

$\textsf{p O H = \frac{1}{2} \left[9.26 - \left(- 1.01090\right)\right]}$

$\textsf{p O H = 5.1354}$

Since $\textsf{p H + p O H = 14}$ we get:

$\textsf{p H = 14 - p O H = 14 - 5.1354 = 8.86}$

As expected, the pH is slightly alkaline.